1) The document is a numerical methods assignment submitted by a student to their professor. It contains 4 problems solved using various numerical methods like Gauss-Seidel iteration, Lagrange interpolation, and Newton divided differences. 2) For problem 1, the solutions to a system of equations using Gauss-Seidel iteration are found to be x1= 1, x2= -2, x3= -3. 3) For problem 4, the value of f(5) is estimated using Newton divided differences to be 92.

1. Numerical Method Assignment
Submitted to Dr. Md. Jashim Uddin
Submitted by Ashikul Islam(173-15-10412)
Section(C)

2. Problem-1:Solve the following system using Gauss-Seidel or Jacobi
iterativeMethod
27𝑥1−2𝑥2−3𝑥3=30
5𝑥1+30𝑥2−8𝑥3=−11
−2𝑥1−2𝑥2+9𝑥3=−19
Solution:
packagedemo;
import java.util.Scanner;
public class DemoTranslation {
public static void main(String[] args) {
int m = 0;
int n = 0;
int l = 0;
float[] x = null;
float[] b = null;
float[] c = null;
float[][] a = null;
final int posLine= 1;
for(intpos = 0; true;) switch(pos) {
default:
c = new float[10];
b = new float[10];

3. x = new float[10];
a = new float[10][10];
System.out.println("nEnter thevalue of n : ");
n = STDIN_SCANNER.nextInt();
System.out.println("nEnter thenumber of iterations : ");
l = STDIN_SCANNER.nextInt();
System.out.println("nEnter theright hand side constants : ");
for(inti = 0; i < n; i++) {
b[i] = STDIN_SCANNER.nextFloat();
}
System.out.println("nEnter thecoefficients row wise: ");
for(inti = 0; i < n; i++) {
x[i] = 0;
for(intj = 0; j < n; j++) {
a[i][j] = STDIN_SCANNER.nextFloat();
}
}
m = 1;
case posLine:
for(inti = 0; i < n; i++) {
c[i] = b[i];
for(intj = 0; j < n; j++) {
if(i != j) {
c[i] = c[i] - a[i][j] * x[j];

4. }
}
}
for(inti = 0; i < n; i++) {
x[i] = c[i] / a[i][i];
}
m++;
if(m <= l) {
pos = posLine;
continue;
} else {
System.out.println("nTheSolution is : ");
for(inti = 0; i < n; i++) {
System.out.printf("nx(%d) =%fn", i, x[i]);
}
}
return;
}
}
public final static Scanner STDIN_SCANNER=new Scanner(System.in);
} }
}

5. So, x1= 1, x2= -2, x3= -3

6. Problem-2:The following values of 𝑥 and 𝑦 are given
𝑥 1.5 5.5 8.5 13.5 15.5
𝑦(𝑥) 4.328 7.097 11.94
4
15.87
5
20.89
6
Find 𝑦(2.5)
Solution:
#include<stdio.h>
#define MAXN100
#define ORDER4
int main()
{
float ax[MAXN+1], ay [MAXN+1], diff[MAXN+1][ORDER+1], nr=1.0,
dr=1.0,x,p,h,yp;
int n,i,j,k;
printf("nEnter the value of n: ");
scanf("%d",&n);
printf("nEnter the values in formx,y:n");
for (i=0;i<=n;i++)
scanf("%f %f",&ax[i],&ay[i]);
printf("nEnter the value of x for which the value of y is wanted: n");
scanf("%f",&x);

7. h=ax[1]-ax[0];
for (i=0;i<=n-1;i++)
diff[i][1] = ay[i+1]-ay[i];
for (j=2;j<=ORDER;j++);
for(i=0;i<=n-j;i++)
diff[i][j] = diff[i+1][j-1] - diff[i][j-1];
i=0;
while (!(ax[i]>x))
i++;
i--;
p = (x-ax[i])/h;
yp = ay[i];
for (k=1;k<=ORDER;k++)
{ nr *=p-k+1;
dr *=k;
yp +=(nr/dr)*diff[i][k];
} printf("nWhen x= %6.1f, corresponding y =%6.2fn",x,yp);
}
So, y(2.5)= 5.27

8. Problem-3:Certain corresponding values of 𝑥 and log10𝑥 are
(11,1.04139),(17,1.23044),(20,1.301029),(23,1.361727)
Estimates the value of log1015 and find the error.
Solution:
packagedemo;
import java.util.Scanner;
public class DemoTranslation {
public static void main(String[] args) {
float[] x = new float[100], y = new float[100];
float a, s = 1, t = 1, k = 0;
int n;
boolean d = true;
System.out.println("nEnter thenumber of the terms of the table: ");
n = STDIN_SCANNER.nextInt();
System.out.println("nEnter therespectivevalues of the variables x
and y: ");
for(inti = 0; i < n; i++) {
x[i] = STDIN_SCANNER.nextFloat();
y[i] = STDIN_SCANNER.nextFloat();
}
System.out.println("nThetableyou entered is as follows :");

9. for(inti = 0; i < n; i++) {
System.out.printf("%.6ft%.6f", x[i], y[i]);
System.out.println();
}
while(d) {
System.out.println(" nEnter the value of the x to find the
respective value of y: ");
a = STDIN_SCANNER.nextFloat();
for(inti = 0; i < n; i++) {
s = 1;
t = 1;
for(intj = 0; j < n; j++) {
if(j != i) {
s = s * (a - x[j]);
t = t * (x[i] - x[j]);
}
}
k = k + (s / t) * y[i];
}
System.out.printf("nTherespectivevalueof the variable y is:
%fn", k);
}
}
public final static Scanner STDIN_SCANNER=new Scanner(System.in);

10. }
So, y(15)= 1.175639

11. Problem-4:Using Newton divided difference formula, find 𝑓(5) from
the followingfunctional values
𝑓(3)=160,𝑓(9)=392,𝑓(13)=952,𝑓(18)=1300,𝑓(23)=1900
Solution:
#include<stdio.h>
#include<conio.h>
void main()
{
int x[10], y[10], p[10];
int k,f,n,i,j=1,f1=1,f2=0;
printf("nEnter the number of observations: n”);
scanf("%d", &n);
printf("nEnter the different values of x:n");
for (i=1;i<=n;i++)
scanf("%d", &x[i]);
printf("nThecorresponding values of y are:n");
for (i=1;i<=n;i++)
scanf("%d", &y[i]);
f=y[1];
printf("nEnter the value of 'k' in f(k) you want to evaluate: n");

12. scanf("%d", &k);
do
{
for (i=1;i<=n-1;i++)
{
p[i] = ((y[i+1]-y[i])/(x[i+j]-x[i]));
y[i]=p[i];
}
f1=1;
for(i=1;i<=j;i++)
{
f1*=(k-x[i]);
}
f2+=(y[1]*f1);
n--;
j++;
}
while(n!=1);
f+=f2;
printf("nf(%d) =%d", k , f);
getch();
}

13. So, f(5)= 92